Problem: Let $a(x)=-2x^{12}+3x^6-x^5+15x$, and $b(x)=x^5$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{-2x^{12}+3x^6-x^5+15x}{x^5}&=\dfrac{-2 {x^{12}}+3 {x^6}- {x^5}}{ {x^5}}+\dfrac{15x}{x^5}\\\\ &={-2x^7+3x-1}+\dfrac{{15x}}{x^5}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${15x}$ is less than the degree of $x^5$, it follows that ${r(x)}={15x}$, and ${q(x)}={-2x^7+3x-1}$. To conclude, $q(x)=-2x^7+3x-1$ $r(x)=15x$ [Is there another way of doing this?]